If the probability of choosing an integer ‘k’ out of 2m integers 1,2,3..2n is inversely proportional to k⁴(1 < k < 2m). If x₁ is the probability that chosen number is odd and x₂ is the probability that chosen number is even, then

Let $\mathrm{P}(\mathrm{k}=\mathrm{i})=\frac{\mathrm{\lambda }}{{\mathrm{k}}^4} \mathrm{for} \mathrm{k}=1,2,3\dots \dots ,2\mathrm{m} \mathrm{where} \mathrm{\lambda } \mathrm{is}$
proportionality constant.
then $\sum \mathrm{P}(\mathrm{k}=\mathrm{i})=1\Rightarrow \mathrm{\lambda }\sum _{\mathrm{k}=1}^{2\mathrm{m}} \left(\frac{1}{{\mathrm{k}}^4}\right)=1$

$$\begin{gathered} \text{ Let }{x}_1\text{ be the probability of choosing the odd number. Then } \\ {x}_1=\sum _{k=1}^m P(2k-1)=\lambda \sum _{k=1}^m \frac{1}{(2k-1{)}^4} \end{gathered}$$

$1-x_1=\sum _{k=1}^m P\left(2k\right)=\lambda \sum _{k=1}^m \frac{1}{(2k{)}^4}<\lambda \sum _{k=1}^m \frac{1}{(2k-1{)}^4}$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1-x_1<x_1\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}_1>1/2\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{x}_2<1/2\end{array}$